Origin of the term "module" for profunctors

Question

Why do they call profunctors "modules"? How do they exactly relate to modules over rings?

Answer 1

It would be more accurate to call them bimodules. Fix a Bénabou cosmos $\mathcal{V}$. Let us say that a $(\mathcal{C}, \mathcal{D})$-bimodule (or profunctor $\mathcal{C} \looparrowright \mathcal{D}$) is a $\mathcal{V}$-functor $\mathcal{D}^\mathrm{op} \otimes \mathcal{C} \to \mathcal{V}$.

What does this mean when $\mathcal{V} = \mathbf{Ab}$ and $\mathcal{C}$ and $\mathcal{D}$ are one-object $\mathcal{V}$-categories? Well, a one-object $\mathbf{Ab}$-category is the same thing as a ring: indeed, given a ring $R$, we can define an $\mathbf{Ab}$-category $\mathcal{B} R$ with a unique object $*$ and $\mathcal{B} R (*, *) = R$ (with the composition corresponding to ring multiplication). And of course a $(\mathcal{C}, \mathcal{D})$-bimodule is literally just that: if $\mathcal{C} = \mathcal{B} R$ and $\mathcal{D} = \mathcal{B} S$, then given an $(R, S)$-bimodule $M$, we define a $(\mathcal{C}, \mathcal{D})$-bimodule $\mathcal{B} M$ by taking $\mathcal{B} M (*, *) = M$, with the obvious action by the "morphisms" of $\mathcal{B} R$ and $\mathcal{B} S$. (After all, what is an $(R, S)$-bimodule if not an abelian group $M$ equipped with a ring homomorphism $S^\mathrm{op} \otimes_\mathbb{Z} R \to \mathrm{End} (M)$?)